If (Sec A + tan A ) (SecB + tanB) (Sec C + tanC) = (Sec A -tanA) (Sec B – tan B) (Sec C – tan C ) . Prove that each of the side is

2 thoughts on “If (Sec A + tan A ) (SecB + tanB) (Sec C + tanC) = (Sec A -tanA) (Sec B – tan B) (Sec C – tan C ) . Prove that each of the side is”

1. Answer:

   ★GIVEN:-

   \sf{(Sec A + tan A ) (SecB + tanB) (Sec C + tanC) = (Sec A -tanA) (Sec B – tan B) (Sec C – tan C ) }(SecA+tanA)(SecB+tanB)(SecC+tanC)=(SecA−tanA)(SecB−tanB)(SecC−tanC)

   ★TO PROVE:-

   each of the side is equal to ±1

   ★FORMULA USED:-

   \sf{Sec^2x – Tan ^2x =1}Sec

   2

   x−Tan

   2

   x=1

   ★LET:-

   (Sec A + tan A ) (SecB + tanB) (Sec C + tanC)——–(1)

   (Sec A -tanA) (Sec B – tan B) (Sec C – tan C )————–(2)

   ★SOLUTION:-

   Multiply Eqs (1) and (2) we get,

   \sf{ (Sec^2A – Tan^2A)(Sec^2B – Tan ^2 B )( Sec ^2 C – Tan ^2 C ) = 1}(Sec

   2

   A−Tan

   2

   A)(Sec

   2

   B−Tan

   2

   B)(Sec

   2

   C−Tan

   2

   C)=1

   \sf{[(Sec A – Tan A)(Sec B – Tan B )(Sec C – Tan C )]^2 = 1}[(SecA−TanA)(SecB−TanB)(SecC−TanC)]

   2

   =1

   \sf{(Sec A – Tan A)(Sec B – Tan B )(Sec C – Tan C ) = \pm \: 1}(SecA−TanA)(SecB−TanB)(SecC−TanC)=±1

   HENCE, each of the side is equal to ±1

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2. Answer:

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